Question

$\frac{6}{3 \sqrt{2} - 2 \sqrt{3} } = 3 \sqrt{2} - a \sqrt{3}$
find the value of a​

Answer 1

GIVEN :-

$\\ \sf \: \dfrac{6}{3 \sqrt{2} - 2 \sqrt{3} } = 3 \sqrt{2} - a \sqrt{3}$

TO FIND :-

• Value of a.

SOLUTION :-

We have ,

$\\ \sf \: \dfrac{6}{3 \sqrt{2} - 2 \sqrt{3} }$

Rationalising the denominator,

Rationalising factor is 3√2 + 2√3.

Multiplying numerator and denominator with (3√2 + 2√3) ,

$\\ \implies \sf \: \dfrac{6}{3 \sqrt{2} - 2 \sqrt{3} } \times \dfrac{3 \sqrt{2} + 2 \sqrt{3} }{3 \sqrt{2} + 2 \sqrt{3} } \\ \\ \\ \implies\sf \: \dfrac{6(3 \sqrt{2} + 2 \sqrt{3}) }{(3 \sqrt{2} - 2 \sqrt{3} )(3 \sqrt{2} + 2 \sqrt{3} ) }$

In denominator,

★ (a-b)(a+b) = a² - b²

• a = 3√2
• b = 2√3

$\\ \implies \sf \: \dfrac{18 \sqrt{2} + 12 \sqrt{3} }{( {3 \sqrt{2}) }^{2} - ( {2 \sqrt{3} )}^{2} } \\ \\ \\ \implies \sf \: \dfrac{18 \sqrt{2} + 12 \sqrt{3} }{18 - 12} \\ \\ \\ \implies \sf \: \dfrac{18 \sqrt{2} + 12 \sqrt{3} }{6}$

$\implies \sf \: \dfrac{18 \sqrt{2} }{6} + \dfrac{12 \sqrt{3} }{6} \\ \\ \\ \implies \underbrace{\sf \: 3 \sqrt{2} + 2 \sqrt{3} }$

This is in the form of 3√2 + a√3 .

Hence , a = 2.